Line l is the bisector of an angle ∠A and B is any point on it. If BP and BQ are
perpendiculars from B to the arms of ∠A, show that B is equidistant from the arms of
∠A.

We are given that line \( l \) is the bisector of \( \angle A \), and point \( B \) is located on this bisector. Perpendiculars \( BP \) and \( BQ \) are drawn from point \( B \) to the two arms of \( \angle A \). We are tasked with proving that \( B \) is equidistant from the arms of \( \angle A \), meaning that the distances \( BP \) and \( BQ \) are equal.

### Step-by-Step Proof:

1. **Label the Points and Distances**:
   Let \( P \) and \( Q \) be the feet of the perpendiculars dropped from \( B \) to the two arms of \( \angle A \). Since \( BP \) and \( BQ \) are perpendiculars to the arms of the angle, \( BP \perp AP \) and \( BQ \perp AQ \), where \( AP \) and \( AQ \) are the two arms of the angle \( \angle A \).

2. **Equal Angles on the Bisector**:
   By definition of an angle bisector, line \( l \) divides \( \angle A \) into two equal parts. Therefore, we have:
   \[
   \angle ABP = \angle ABQ
   \]
   Both of these angles are equal since \( l \) bisects \( \angle A \), and both \( P \) and \( Q \) are projections of \( B \) on the two arms of the angle.

3. **Use of Right Triangles**:
   The triangles \( \triangle BPB \) and \( \triangle BQB \) are right triangles because \( BP \perp AP \) and \( BQ \perp AQ \).

4. **Congruent Triangles**:
   Now consider the triangles \( \triangle BAP \) and \( \triangle BAQ \):
   - Both triangles share the side \( AB \), as \( B \) is on the bisector.
   - \( \angle ABP = \angle ABQ \), as explained earlier.
   - The angles \( \angle BPA = \angle BQA = 90^\circ \) because \( BP \) and \( BQ \) are perpendiculars.

   Hence, by the **AAS (Angle-Angle-Side)** criterion, the triangles \( \triangle BAP \) and \( \triangle BAQ

Line l is the bisector of an angle ∠A and B is any point on it. If BP and BQ are perpendiculars from B to the arms of ∠A, show that B is equidistant from the arms of ∠A.

This geometry problem involves proving that point B, located on the bisector of ∠A, is equidistant from the arms of the angle. By using the angle bisector theorem and triangle congruence (AAS), this step-by-step solution is ideal for students in Grades 9-10.

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